Is $f_n(x_n) \to f(x)$ Given $f_n \to f$ Uniformly and $x_n \to x$?

[evinda]

Hey! :)
Let $f_n: \mathbb{R} \to \mathbb{R}$ continuous,such that $f_n \to f$ uniformly.Also,let $x_n$ a sequence of real numbers,such that $x_n \to x$.Show that $f_n(x_n) \to f(x).$

Since $f_n$ continuous and $f_n \to f$ uniformly,we know that $f$ is continuous and that:
$ \text{ sup } \{ f_n(x)-f(x): x \in \mathbb{R} \} \to 0 $
As $x_n \to x: \forall \epsilon>0 \exists n_0 \text{ such that } \forall n \geq n_0: |x_n-x|< \epsilon$.

Can I conclude from the last relation that $f(x_n) \to f(x)$ ? If yes,how? Do I take $x_0=x_n$ at the relation $|x-x_0|< \delta \Rightarrow |f(x)-f(x_0)|< \epsilon$ ? (Thinking)

 

[Opalg]

Given $\varepsilon>0$, the uniform convergence of the sequence $(f_n)$ tells you that there exists $M$ such that $n>M\;\Rightarrow\;|f_n(t) - f(t)| < \varepsilon$ for all $t$ in $\mathbb{R}$. The continuity of $f$ tells you that there exists $N$ such that $n>N\;\Rightarrow\;|f(x_n) - f(x)| < \varepsilon$. You want both those things to be true. So let $n_0 = \max\{M,N\}.$ Then use the triangle inequality to conclude that $$n>n_0\;\Rightarrow\;|f_n(x_n) - f(x)| \leqslant |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| < \varepsilon + \varepsilon.$$

 

[evinda]

The continuity of $f$ tells you that there exists $N$ such that $n>N\;\Rightarrow\;|f(x_n) - f(x)| < \varepsilon$.

How did you get to the relation $|f(x_n) - f(x)| < \varepsilon$ by the continuity of $f$?